lib/expr: use the IX register a bit less

It's an awkward register to use and avoiding its use allows us to strip the
resulting binary significantly. parseEXPR keeps the same signature though.
This commit is contained in:
Virgil Dupras 2019-11-22 19:56:08 -05:00
parent 2f71ad6d2f
commit 501fe96e07

View File

@ -1,6 +1,7 @@
; *** Requirements ***
; findchar
; multDEBC
; callIXI
;
; *** Defines ***
;
@ -14,6 +15,9 @@
; **This routine mutates (HL).**
; We expect (HL) to be disposable: we mutate it to avoid having to make a copy.
; Sets Z on success, unset on error.
; TODO: the IX output register is a bit awkward. Nearly everywhere, I need
; to push \ pop that thing. See if we could return the result in DE
; instead.
parseExpr:
push de
push hl
@ -38,15 +42,16 @@ _parseExpr:
; Operator found, string splitted. Left in (HL), right in (DE)
call _resolveLeftAndRight
; Whether _resolveLeftAndRight was a success, we pop our lvl 1 stack
; out, which contains our operator row. We pop it in HL because we
; don't need our string anymore. L-R numbers are parsed, and in DE and
; IX.
pop hl ; <-- lvl 1
; out, which contains our operator row. We pop it in IX.
; L-R numbers are parsed in HL (left) and DE (right).
pop ix ; <-- lvl 1
ret nz
; Resolving left and right succeeded, proceed!
inc hl ; point to routine pointer
call intoHL
jp (hl)
inc ix ; point to routine pointer
call callIXI
push de \ pop ix
cp a ; ensure Z
ret
; Given a string in (HL) and a separator char in A, return a splitted string,
; that is, the same (HL) string but with the found A char replaced by a null
@ -90,7 +95,7 @@ _findAndSplit:
.find:
; parse expression on the left (HL) and the right (DE) and put the results in
; DE (left) and IX (right)
; HL (left) and DE (right)
_resolveLeftAndRight:
call parseExpr
ret nz ; return immediately if error
@ -98,12 +103,14 @@ _resolveLeftAndRight:
; IX. What we need to do now is the same thing on (DE) and then apply
; the + operator. Let's save IX somewhere and parse this.
ex de, hl ; right expr now in HL
push ix
pop de ; numeric left expr result in DE
jp parseExpr
push ix ; --> lvl 1
call parseExpr
pop hl ; <-- lvl 1. left
push ix \ pop de ; right
ret ; Z is parseExpr's result
; Routines in here all have the same signature: they take two numbers, DE (left)
; and IX (right), apply the operator and put the resulting number in IX.
; and IX (right), apply the operator and put the resulting number in DE.
; The table has 3 bytes per row: 1 byte for operator and 2 bytes for routine
; pointer.
exprTbl:
@ -130,107 +137,87 @@ exprTbl:
.db 0 ; end of table
.plus:
add ix, de
cp a ; ensure Z
add hl, de
ex de, hl
ret
.minus:
push ix
pop hl
ex de, hl
scf \ ccf
or a ; clear carry
sbc hl, de
push hl
pop ix
cp a ; ensure Z
ex de, hl
ret
.mult:
push ix \ pop bc
call multDEBC
push hl \ pop ix
cp a ; ensure Z
ld b, h
ld c, l
call multDEBC ; --> HL
ex de, hl
ret
.div:
; divide takes HL/DE
push bc
ex de, hl
push ix \ pop de
call divide
push bc \ pop ix
ld e, c
ld d, b
pop bc
cp a ; ensure Z
ret
.mod:
call .div
push hl \ pop ix
ex de, hl
ret
.and:
push ix \ pop hl
ld a, h
and d
ld h, a
ld d, a
ld a, l
and e
ld l, a
push hl \ pop ix
cp a ; ensure Z
ld e, a
ret
.or:
push ix \ pop hl
ld a, h
or d
ld h, a
ld d, a
ld a, l
or e
ld l, a
push hl \ pop ix
cp a ; ensure Z
ld e, a
ret
.xor:
push ix \ pop hl
ld a, h
xor d
ld h, a
ld d, a
ld a, l
xor e
ld l, a
push hl \ pop ix
cp a ; ensure Z
ld e, a
ret
.rshift:
push ix \ pop hl
ld a, l
ld a, e
and 0xf
ret z
push bc
ld b, a
.rshiftLoop:
srl d
rr e
srl h
rr l
djnz .rshiftLoop
push de \ pop ix
ex de, hl
pop bc
cp a ; ensure Z
ret
.lshift:
push ix \ pop hl
ld a, l
ld a, e
and 0xf
ret z
push bc
ld b, a
.lshiftLoop:
sla e
rl d
sla l
rl h
djnz .lshiftLoop
push de \ pop ix
ex de, hl
pop bc
cp a ; ensure Z
ret